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Immobile pressure-sensitive stick
I was reading the Japanese sci-fi novel Yukikaze (greatly recommended btw) again and I came across a detail that describes that the stick of the fighter plane the main character is flying is fixed in place and does not tilt around like those in a normal airplane, and instead it can detect the pressure exerted on it by the pilot changing their grip on the stick as they flew the plane. It claims that "under the correct conditions it could execute a pilot’s will with lightning speed." Would this design actually be viable and useful? The book does describe an issue where after sustaining an injury that causes the pilot to lose most feeling in his right arm he puts too much pressure into the stick without realizing it and caused the plane to fly unstably, but I imagine this would still happen with a normal control stick if you couldn't feel your arm.

This is an actual design, used in the early iterations of the F-16 fighter jet. The ideology behind the design was much the same you described, with the additions on the "solid" stick mitigating the effects of G-forces on controlling the plane.
The design received much critique from the pilots, eventually leading to re-designing the stick such that it had some movement in it.
https://en.wikipedia.org/wiki/General_Dynamics_F-16_Fighting_Falcon#Cockpit_and_ergonomics

Oh I never knew they actually made one in real life, that's really cool. Is there any place I can find out more about this one that was fitted on the F-16?

It was a pretty horrible idea actually, and USAF test pilots agreed.  It makes as much sense as a stationary steering wheel in your car, or fixed handlebars on a bicycle.

@user721108, If there's a lot of slop in the controls then that's a problem.  But, easily taken up, and not worth the massive compromise of an immovable input device.  Humans need tactile feedback to control rate and magnitude.

@user721108, but what purpose would shortening the travel path accomplish? Flight controls and steering wheels both become more sensitive at speed, naturally, without any help or extra complexity. –

The rigid stick also took away the 'feel' of the aircraft.  In a normal aircraft with pilot controls connected to the control surfaces, you end up with both deflections and stick forces that are nearly linear with g-forces.  I.e. stick angle and stick force per g are meaningful quantities.  A pilot can feel how far they are deflecting and they can feel how much the control surface is pushing back.  Most fly by wire systems work hard to duplicate this feel.  The rigid F-16 system actually lead to pilot injuries.  They would blow out their tendons pulling so hard.

The YF-16 and the F-16 use the same Side Stick Controller. Both measure displacement against a spring force to “sense” the pilot’s input forces. The F-16 has larger displacements than the YF, but they both require displacement.


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Why are bridles no longer used on carriers?
So, back in the day a lot of carriers had a "bridle" fitted onto the catapult. This was basically a V-shaped cable that attached to two points on the underside of a plane and would basically secure it to the catapult shuttle as it launched it down the deck. This configuration helps with launching the planes better since it is connected to the center of mass from what I've heard, particularly useful when FAA Phantoms had to launch from the smaller deck of the HMS Ark Royal. However, this system I have only seen on older carrier jets (your F-4s, A-4s, Etendards, etc) and all the modern ones lock the nosewheel to the shuttle, in fact several modern carriers still have this extra lip on the deck where the catapult would continue on after the plane disconnected and stop there so the bridle could be recovered without flinging it out into the sea. What's the reason for its replacement?

Somewhat related -- https://aviation.stackexchange.com/a/65857/34686

@BryBuriya-- I created a chat room for this question; welcome to join in-- https://chat.stackexchange.com/rooms/143281/a-room-to-talk-about-catapult-launch-dynamics

Grumman developed the nosewheel launch bar system for the A-6, E-2 and C-2 aircraft back in the 1960s. And they quickly caught on for several reasons:

Greater simplicity.  The bridle arrangement has multiple moving parts that attached to the aircraft, and once launched, must be either recovered, or allowed to be jettisoned overboard. Bridles were equipped with bungees to prevent them from going overboard.  In addition to the design of the hydraulic holdback bars in the 1970s, it reduce the risk of FOD associated with the frangible holdback (dog bone) design.

Safety.  Aircraft carrier flight decks are dangerous places to work even in the best of times. Add into this the problems associated with the launch bridles, and they become even more hazardous.  The bridles and the holdback pendants have to be manually installed on the aircraft after it taxis over the catapult by deck personnel.  This puts more squishy, pink meat sacks in range of very dangerous things like hot jet blast, jet intakes, propellers, flight control surfaces etc. all while handling heavy steel ropes.  Personnel have to remain in place to make sure the bridle remains secured to the aircraft until the catapult can be placed in tension prior to launch.  Should a bridle fail during the launch stroke, it can whip free of the catapult shuttle, and go flying off down the deck in an uncontrolled fashion potentially either seriously injuring or killing aircraft handlers and/or causing damage to the ship or aircraft parked nearby.  As an aircraft leaves the flight deck just after launch, the free ends of a bridle potentially could strike the aircraft underside, causing damage.  Then there’s the problem of having this wire rope, flung over the deck at 150 mph, which then needs to be rested if you want to recover it. If anything goes wrong in that process, it becomes very dangerous to personnel and equipment on the ship.

Ease of operations increase launch tempos.  If it’s easier to connect a jet to the catapult, and shoot it, it takes less time to launch it. This increases the launch tempo capabilities of the ship, getting more aircraft out of the sky, quicker than if the airplanes require the use of a launch bridal to shoot.

WAY better answer than mine, thanks!  (love "squishy pink meat sacks...")

"Hot jet blast" indeed. While it was not jet blast (thankfully) I've been burned on the exhaust of just regular engines.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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What are the chances of a fresh pilot school graduate to get a 'trainee' position at an airline in Europe?
My daughter (19) is about to start a career as a pilot. She completed the entry test to a pilot school in Norway (we live there) and got her Class 1 medical license. She's going to enroll for a 2-year training program, where - according to the school's training plan - she'll end up with ca. 230 logged hours (flight + sim combined) and an EASA CPL(A) IR ME with frozen ATPL(A) plus MCC.
Financing these two years is a significant investment, therefore I'd like to be sure that this really takes her somewhere with opportunities to continue. I looked around on some pilot career portals and career pages of major European airlines and I found that the qualifications listed above are roughly in line with the minimum requirements of entry programs at some airlines. My impression is that some job opportunities do exist for candidates with that level of (in)experience, but I'm not able to find any numbers e.g. for the number of positions open vs. the number of fresh pilots graduating from schools each year.
After completing that 2-year program, will my daughter stand in line as candidate nr. 1000 for a beginner's position at an airline? Or do airlines have such a high demand for new pilots that virtually anyone hitting the bare minimum can get in?

You should probably start by asking questions to the school offering the training program like "What percentage of your graduates are able to find a job as an airline pilot immediately upon graduation?"  Also, "What is the average starting pay for your graduates?"  Things like that will give you some practical insight into the "value" of their program.  If they don't have good answers, then the answers are probably not good.

As far as I'm aware, there are also pilot schools directly from airlines, such as Lufthansa and Condor. But I don't know their pricing and their accessibility in Norway. Graduating from one of those would probably profit the opportunities after graduation.

There are many pilot forums online which are a better place to ask this kind of question.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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How many master cylinders are there in a typical dual brakes Cessna 172?
After searching in the AFMs of the several C172 my flight school operates, all of them describe the system very basically with no diagrams, as exemplified in the following picture:

Is there one master cylinder per pedal - making the total 4 master cylinders?
Or are the pilot and copilot pedals mechanically linked and each side connected to a shared pressure cylinder (making the total 2, one for each main wheel)?
Note I am not asking about the particulars of a specific model, but about the more general configuration one could expect to find in most C172s.

2 master cylinders.  They are on the pilot's rudder pedals.  The pilot and co-pilots rudder pedals are mechanically linked so that pressing either of the co-pilot's rudder pedals moves the pilot's rudder pedals and vice versa.  In either case, moving the rudder pedals actuates the master cylinder for that brake line.
The POH is specific in its wording.  It does not say, "a master cylinder attached to each of the pilot's and co-pilots rudder pedals."
Some models do not have co-pilot's toe-brakes.
To quote the maintenance manual:
"5-59. DESCRIPTION. The brake master cylinders, located immediately forward of the pilot's rudder pedals, are actuated by applying pressure at the top of the rudder pedals. A small reservoir is incorporated into each master cylinder for the fluid supply.  When dual brakes are installed, mechanical linkage permits the copilot pedals to operate the cylinders."

Thanks - nice work with the edit!


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How do I calculate induced drag?
How would I go about solving this problem?
An aircraft has 6000lbs of total lift and a downwash angle of 6 degrees. Remember that the total lift vector is tilted aft by HALF the downwash angle. What is your induced drag?

what have you tried? why do you think you don't know how to answer?

You would solve this problem by drawing the lift vector being tilted back at half the downwash angle, then apply some trigonometry to find the horizontal bit of the lift vector.


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Why do many small piston singles have panel-mount engine controls?
Most small single engine piston aircraft (C152, C172, C182, Pipers, etc) have panel-mount controls for throttle, mixture and prop, which sit quite far forward compared to where it would be comfortable to rest one’s hand, such that a centre-console quadrant would appear to be much more comfortable.
Why don’t more small aircraft place controls in this apparently superior position?

That's where they fit! On both the original 172 and the Piper designs the "center console" area is where the flap bar is mounted, thus preventing the throttle from going there. The C172 has since moved to electric flaps but did not relocate the throttle in the process. Piper has retained that flap bar on its PA-28 series it still produces to this day. Many of the PA-28's also have their elevator trim wheel in that area (older ones have overhead trim)
It should be noted that new airframes designed in the last two decades like the SR-2X and the Diamonds do have throttles mounted as you discuss.
Piston Pipers also lack center consoles in general as they only have one door and the pilot would have to climb over a center console to get in if one were in place.
FWIW the Piper Cub (designed before the PA-28's and 172) have the throttle right where your left hand sits.

On the panel would also seem cheaper to build and more reliable, given a shorter & straighter cable run.  And really, what percent of the time do you want to rest your hand on throttle &c?  Not having important stuff that can be knocked out of position when you move around also makes for more leg & seat room.


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Hypersonic Boost Glide Range
Looking at an article earlier about hypersonic boost glide vehicles and I noticed how small their L/D ratios were ~2.5/1.  If they begin their hypersonic glide at 50km in altitude would they not travel forward 2.5km for every km they lose?  Hence their range would be 50km * 2.5 = 125km, clearly this isn't correct and their published ranges are 6,000+km.  What equations would one use to calculate their range since its not simply a matter of L/D.

At high speeds the centripetal force cannot be neglected. Low earth orbit satellites move at 7.8 km/s and do not need lift at all to stay at the same altitude.

Gliding implies that lift equals weight.
This is not the case for hypersonic boost glide vehicles. They ascend like rockets during the burn phase (or are carried aloft by a rocket) and then fly mostly, but not fully, ballistic after their engine has shut down (rsp. have been released from their carrier). In contrast to a conventional reentry vehicle, they are designed to create aerodynamic forces perpendicular to their flight path for trajectory modification. This allows them to boost their range a bit, hence the name, which is more efficient than creating the same force with a rocket. However, the major reason for their use of aerodynamic steering is to make their final trajectory harder to predict.
Hypersonic soaring was initially proposed by Eugen Sänger to stretch the range of a hypersonic rocket plane. By "dipping" into the denser atmosphere and creating enough lift for a trajectory change up into near-space, it was hoped to fly halfway around the Earth. The US research program that lead to the Boeing X-20 "Dyna Soar" was an application of this idea. However, the heat loads in the lift phase are too high for current materials. Only instationary hypersonic flight with either ablative coatings or heat sinks is so far possible.

When the engine stops, the craft is still travelling upwards at a steep angle. In the thin atmosphere at 50 km, its trajectory is more or less ballistic and takes a while to reach its zenith, turn back down and enter thicker air where its glide ratio takes over. During that time it has travelled a long way.
It can then trade its high speed for extra lift, by increasing its AoA and effectively trying to climb back out, slowing itself down in the process. The poor gliding angle is then relative to that "virtual" climb flightpath.

From a congressional briefing on the subject:

Hypersonic glide vehicles (HGVs), like all weapons delivered by medium- and longer-range rocket boosters, can travel at speeds of at least Mach 5, or about 1 mile per second.The key difference between missiles armed with HGVs and missiles armed with ballistic reentry vehicles (i.e., those that travel on a ballistic trajectory throughout their flight) is not their speed, but their ability to maneuver and change course after they are released from their rocket boosters.

So yes, the hypersonic glider does have a worse L/D ratio than an aeroplane, still better than a ballistic missile though.

One may wonder how well a 1 mile/second object can turn, especially in thin air.

This is a very interesting question from the other end of the gliding spectrum.
We must consider the two friends of gliders, altitude (potential) and speed (kinetic) energy.
An incoming hypersonic vehicle is blessed with both.  Much of its low lift to drag ratio is due to its very high velocity.  So it would glide more effectively by trading speed, rather than altitude, as "fuel" against drag at first, then use its wing and altitude for maximum range.  This was the approach the Space Shuttle used to re-enter and land from thousands of miles away.
For military purposes, obviously it wants to keep its speed for as long as possible to avoid interception.
The end design is applications driven.

After some further suggested digging (this question was cross-posted on Reddit) I came across the following equation for general hypersonic boost glide gliding ranges, its in this paper on page 10:  link


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What is this high-wing single-engine plane performing a short landing?

Source: Shortest Landing!!! Severe Headwind! Aircraft hovers and vertically touches down! (YouTube)
What is this high-wing single-engine plane performing a short landing?

looks like a Piper Cub or one of its million derivatives. Picture doesn't show enough detail to be much more precise.

It looks more like a PA-18 super cub than a J-3 cub to me because it's slightly beefier, but it's very hard to tell the difference from that angle.

A J-3 has the front windshield struts intersect at the top at center in an inverted V. The Supercub is the other way around.  That's a Supercub.

Much more detail is visible in this animated gif of it apparently taking off backwards in a strong headwind. (Sorry, stack.imgur.com can't understand this image format.) https://cs9.pikabu.ru/images/big_size_comm_an/2017-06_2/1496817602161265640.gif

The video description (thanks to ymb1 for adding the link) says:

By Jon Bush.

If you search for that name and STOL, you will find his YouTube channel, where you can find a higher quality video of that landing:

(Ultimate Alaskan Plane Landing, Zero Feet!)
The video description says:

Having too much fun flying this Experimental Stol Quest SQ12 supercub  in the wind on the Knik River in Alaska @akbushpilot.

So the aircraft is an STOL Quest SQ12 Super Cub, an experimental plane based on the design of the Piper PA-18 Super Cub.

Thank you so much

This is a Piper Cub which has been modified with a high-performance propeller and tundra tires. It probably also has either a 150 or 180 HP engine in it.


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What is this commercial aircraft based on its cockpit?

So I saw a meme featuring this photo on Reddit (although the text makes me think its from TikTok) and I was wondering what the aircraft was. It has a yoke, so its not Airbus, and the show the pilot is watching is Eromanga Sensei, which means this photo was taken in April 9, 2017 or later, although it does appear to have a somewhat old cockpit. It also appears to have what looks like eyebrow windows, although I'm probably wrong about that. I also have a higher quality but slightly photoshopped image.

I also noticed that the pilot is wearing what looks like an oxygen mask, and the nose of the aircraft seems to be abruptly cut off, so could this be a simulator? Thanks!

FWIW, the background appears to include a dumpster, the back of a semi trailer, and a jetway. Not to mention that if it were a sim with a cockpit that has windows, there'd be a wrap-around display screen outside. So seems very unlikely this is a simulator.

The watermark says @ma.tter so maybe you can try and hit them up on any of the socials?

As for pilot apparently playing or watching something on a tablet, my guess would be he just learned they'll be an hour delayed due to flow control or just sent out for a mechanic to fix something that didn't pass the pre-flight checks and has therefore a lot of time to kill right now.

How certain are you the content on the iPad is not edited in? If it is (which is my guess) then it has no bearing on the date of the original photo. By the way, this image has been circulating on Reddit (and other sites) since before TikTok was a thing... and the text is from Snapchat anyway, not TikTok

@randomhead I agree... I would expect the lower quality image to be the photoshopped version, if for no other reason than this makes it harder to detect the alterations.

Oh man, if I ever write electronic flight bag software, I an so making all the screens look like computer games...

That is a McDonnell Douglas MD-80. You can clearly see the characteristic engine fire handle for the left engine in your second picture (red circle, the one for the right engine is hidden behind the oxygen tube):

The picture also shows the first officer's Primary Flight Display (blue circle). These were still analog instruments in the Douglas DC-9 and earlier MD-80 variants. Here is a view of an MD-80 cockpit for reference:

(source: YouTube: MD80 Cockpit Takeoff Part 2 FULL HD)

Any comment on why the pilot appears to be using an oxygen mask?

@MichaelMacAskill I think that is just an optical illusion. Unlike more modern airliners, where the oxygen masks are hidden in the side panels, the MD-80 has them hanging from the ceiling next to the pilots (see e.g. this YouTube video). If the pilot would wear the oxygen mask, the straps would be around the ear.

@MichaelMacAskill If this photo was taken in the last year and a half or so, I'd think it's pretty obvious why the pilot is wearing a mask...

@DarrelHoffman Yes, but the meme already existed in 2018, so the picture is pre-COVID.

It looks like a DC9, or an MD 80 by the cockpit windows.


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wave drag: Lift coefficient in Korn's Equation
I have been using the Korn equation:
$$
M_c = \frac{0.95}{cos(\lambda)}-\frac{t/c}{cos(\lambda)^2}-\frac{C_L}{10cos(\lambda)^3}-\sqrt[3]{\frac{0.1}{80}}
$$
to estimate the critical mach number of a swept wing, but I am now suddenly doubting the 3rd term.
The lift coefficient over a swept airfoil at transonic speeds is lower than incompressible theory would suggest, and there are various corrections used to allow you to use incompressible theory.
My question is, is that taken into account in the Korn equation above, such that I can just do the usual $C_L = \frac{2L}{SM^2p\gamma}$ or do I need to use a corrected lift coefficient?
N.B. the Korn equation was for airfoils wasn't it, and I am trying to use it for an entire wing...is this whole usage flawed?

Korn never wrote the Korn equation.
Korn made the observation that the critical Mach number of airfoils displayed a linear relationship with t/c and design lift coefficient.
Others introduced the airfoil technology factor (the 0.95 here) and corrections for application to 3D wings.  The cube root of 0.1/80 appears to be a correction between drag divergence and critical Mach numbers.
So, when you're using a complex 'Korn Equation' like this, you really have to follow your references back to who derived it.  There is a bunch of related work that likely leads to Mason or Takahashi or a few others.
In any case, make sure you have a good source for your 'Korn Equation' and then use it the way the author describes in their paper.

Thanks Rob, yeah that's kind of the issue, I can't find a source for this one...Guess I'll keep looking, I'll check out Mason and Takahashi!

If you're using an equation from a source that either doesn't cite where it comes from -- or very clearly explain everything about it -- it probably isn't a good source.  Check out this reference https://archive.aoe.vt.edu/mason/Mason_f/ConfigAeroTransonics.pdf Note also that the user of upper case Lambda $\Lambda$ is for sweep, where lower case lambda $\lambda$ is often used for taper ratio.


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Logging instrument time with a simulator with only a CFI present
I am a private pilot working on my instrument rating.  I have all my required instruction time and am working to build time to my 40 hours.  My new flight school has a qualified simulator. My CFII is on long term leave. Can I use the simulator with just a CFI and log simulated time?  The Federal Aviation Regulation, 14 CFR Part 61.65 (h), just says authorized instructor.

Thanks for the correction…yes, authorized. Its subsection h under 61.65

FAR 61.65(d)(2) is a little more nuanced than a simple yes or no answer.

Forty hours of actual or simulated instrument time in the areas of operation listed in paragraph (c) of this section, of which 15 hours must have been received from an authorized instructor who holds an instrument-airplane rating, and the instrument time includes:

The way I read this is a CFI (no II) can provide instruction for 25 of the required 40 hours.  A CFII must provide at least 15 hours.
Do I think this is wise?  No!
An authorized instructor (CFI or CFII) is required for simulator time to count towards an instrument rating.

My interpretation of https://www.ecfr.gov/current/title-14/chapter-I/subchapter-D/part-61#p-61.65(h) is that this time must be logged under the administration of a authorized flight instructor who has an instrument rating on his/her flight instructor certificate.  So, no, you may not do this on your own  and log the time toward an instrument rating without the supervision of a CFII present and signed off as dual instruction time.  This isn’t to say that you cannot practice flying the simulator on instruments on your own, but you will not be able to log that time toward your rating.

Once the OP gets his or her instrument rating then they could log this approved simulator time towards IR currency without any instructor at all.
So OP should stay in good graces with the sim's owner.


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Crossing mid-field on a circling approach
During a circling approach, can I cross midfield to enter the pattern or do I have to literally fly around the airfield to get into the pattern?  I know its an odd question, but my airfield has an RP on one runaway and it would feel safer to cross midfield and enter the pattern

Pardon my ignorance, but what is an RP?

During an IFR Circling approach, in IFR conditions, there is no pattern. You own the airspace and can fly anywhere you want, unless it is restricted by a published circling procedure. (example: no circling south of the airport)
In VFR conditions, there might be VFR traffic to contend with. The tower or local VFR procedures would then come into play.

I agree, but I will complicate the issue somewhat. If you break out of the clouds and the airport has VFR conditions AND VFR traffic in the pattern, then the circling approach is expected to join the VFR traffic in a normal pattern entry. This could happen with a marginal VFR ceiling, depending on the circling minimum and such, though VFR traffic is usually rarified on those days.

Thanks, I edited my answer to reflect possible VFR conditions.

@MikeSowsun I did circling apprs all the time in the CRJ sim but never tried a real one. The standard missed approach/engine failure/fire demo for training and check rides was usually done from a circling approach just as you were getting lined up on short final, dialing the stress level to 11. With the US Regionals, circling maneuvers were forbidden by most operators.


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Request 2(both) engine startup then pushback at the gate?
Usually, the engine starts after pushback.
but if the pilot requsests two(both) engines startup before pushback, can the controller approve it?

This can be done, and can be approved when needed.
The most common case would be an inoperative APU, requiring the use of external high pressure air to start an engine. Once one engine is started, you can push back and perform a "crossbleed" start for the other engine, using bleed air from the operating engine to start the other one. That does require power above idle on the operating engine, and sometimes it's easier to just use the external air to start both engines at the gate before pushback.
It's important to keep vehicles, especially large, high profile vehicles like catering trucks, away from the tail while the engines are running; that's accomplished with marshallers with lighted wands. ATC would protect the area behind the aircraft from light aircraft for the same reason; on a big airline ramp, that isn't so much concern, but at a smaller airport with a taxiway close behind the gate, it could be.
So you tell Ground Control what you need to do, they wait for (or create) a break in traffic flow behind you, and then they approve it. And when the engines are started, you request clearance for pushback as normal, and go on with your day.

Ramp space is usually not controlled by ATC.

@RetiredATC Technically not, but Ramp Control (when operating) functions as such and issues startup/pushback clearances the same way that Ground Control would.  In some contexts, the distinction between actual ATC and Ramp Control matters, but when you're in the cockpit requesting clearance to push back or start engines at the gate, the distinction is basically lost.

Is ramp control regulatory or advisory? If a pilot pushed back w/o ramp control permission, could a Pilot Deviation be filed?

Reality: get permission from Ramp Control before you push (assuming they're operating & issuing those). Can they file a PD? AFAIK, no, but that doesn't change the fact that, de facto, they do control movement on the ramp. (Besides, a pilot who doesn't follow the published procedures for getting permission from Ramp opens himself up to delays, or in the bad case, "careless & reckless" possibilities.)


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Spanwise lift distribution units
I was recently reading NASA/Boeing’s report on the Transonic Truss Braced Wing (the SUGAR IV final report), and noticed that when they graph the spanload for the design they graph lift coefficient multiplied by normalised chord against span, rather than just lift coefficient against span. I guess this is a pretty good proxy for circulation minus a few constants.
My question is, I’ve seen this described elsewhere as a ‘load distribution’ rather than a ‘lift distribution’, what is the difference, and if I were to graph lift coefficient against span would this not be a spanload (unless I have a rectangular wing) since it isn’t a good expression of circulation?

2D (sectional) lift coefficient is usually abbreviated $c_l$, and the sectional lift is usually $l$.  The former is dimensionless, the latter has units of force/length.
$l=c_l\,q\,c$
For 3D, we use $C_L$ and $L$
$L=C_L\,q\,S$
When integrating from root to tip along a wing, we can write
$S=2*\int_0^{b/2} c(y)\,dy$
Where the 2 is to capture both halves of the wing, and $b$ is the wingspan.  I have written $c(y)$ to emphasize that the chord is a function of the span -- i.e. the wing is tapered.
We might then define $\bar{c}$ such that:
$S=b\,\bar{c}$
We can also write (for a symmetrical condition)
$L=2*\int_0^{b/2} l(y)\,dy$
Here $l(y)$ is the load distribution.  The distribution of sectional force that we integrate to get the total lift.  This is what matters for induced drag minimization.
Substituting the definition of $l$ above (and pulling out the dynamic pressure)...
$L=q\,2*\int_0^{b/2} c_l\,c(y)\,dy$
Here you can see where the practice of plotting $c_l\,c$ comes from.  It is proportional to the local load distribution.
Applying the definition of $C_L$
$C_L=\frac{2}{S}*\int_0^{b/2} c_l\,c(y)\,dy$
Where $S$ can be written...
$C_L=\frac{2}{b\,\bar{c}}*\int_0^{b/2} c_l\,c(y)\,dy$
If you pull the $\bar{c}$ back in, you get
$C_L=\frac{2}{b}*\int_0^{b/2} \frac{c_l\,c(y)}{\bar{c}}dy$
Which shows us that this term:
$\frac{c_l\,c(y)}{\bar{c}}$
Is proportional to the load distribution, but it is scaled to be similar to lift coefficient.
When you compare plotting the above with just $c_l$, you see that the difference is in the weighting by the local chord.
Consider a wing with a lot of taper.  Imagine that it somehow has the same lift coefficient everywhere.  The root section with large chord contributes more to the load than does the tip section with a short chord.
Both plots are useful.
$\frac{c_l\,c(y)}{\bar{c}}$
Is good for comparing to an elliptical shape to understand how good the induced drag will be for the wing.  It is also important for the starting point of structural calculations.
$c_l(y)$
Is good for considering how 'close' you are to stall.  It shows how 'hard' the local airfoil is working.
A wing with uniform load distribution ($l(y)$), but with lots of taper will have a $c_l(y)$ that increases sharply at the tips.  That is a wing that is prone to tip stall.

Thanks! So if I want a wing to follow a particular lift distribution for drag minimisation purposes, I need to plot c_l(y)*c(y)/cbar rather than just purely cl(y).

@AlastairWyllie That is correct.  They will look the same if c(y) is constant, but otherwise, you need to keep those in there.


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Circulation around an airfoil at negative α
I was recently reading this very helpful string What is the physical meaning of circulation found in Kutta condition? and it explained very nicely potential flow around an airfoil. My question then is this: what happens to circulation around a cambered airfoil at a small, negative angle of attack which is producing lift, but is also producing upwash. How do we reconcile Kutta-Жуковский theorem with this observation?

For lift to exist, a net downwash has to be present too, because action and reaction are opposite and equal...

Well okay sure. But Lift is equal to ρvΓl where gamma is circulation, rho is density, l is length (of chord) and v is of course velocity. Impulse accounts for only half this and is caused by the turning of the flow (which is caused by adding the circulation into a streamline mathematical model). So yes. But also. As per A. Bowers et al (in ‘On Wings of the Minimum induced drag: Spanload implications for aircraft and birds’) such airfoils do exist in nature and do produce lift if very little.

You can't have lift without downwash. It's a physical impossibility...

There is always upwash and downwash. The only impact of a negative attack angle is that the circulation will be opposite and the lift will go downwards instead of upwards.

If circulation is opposite it means г is negative and lift is not produced. Lift IS produced no matter what is said by xxavier. I agree with them  we have no idea how but I am certain that Al Bowers is correct and further experiments by amateurs confirm his experimental, and analytical evidence. There has to be positive circulation or a new nuance to the theorem of lift.

Why should lift be zero if $\Gamma$ is negative? If $\Gamma$ is negative then lift is negative i.e. pointing downward, that's all.

Yes. Sorry, that’s what I meant. I don’t consider it lift if it’s acting down, it’s downforce. For those talking about downwash, think about cambered airfoils, some of them have a zero lift angle at around -5°, so if you held one at -3°, you would have lift but negative AoA. The question is concerning when |AoA|>|downwash angle| so that the negative AoA overpowers the downwash angle, note, there is a downwash angle, so lift is still produced

Thanks to everyone who has posted answers and comments on this over the years.
My question has an incorrect premise in it which was pointed out by some people and my responses reveal some sloppy use of terminology. ‘Upwash’ is probably better thought of as created by a wing, in that respect, in the wake of any airfoil producing lift there must be a downwash component, or else lift is not being produced.
However, in my mind I was considering that this component was not sufficient to turn the incident airflow from an upward to a downward direction. That the airflow is at such an angle should have been made clear in the question, but was not because I did not realise this at the time. As can be seen in some of the discussion, the reason this is of importance is that the only way this is possible is if some other effect is causing that updraft, perhaps a nearby body or another wing, or 3D effects from a different part of the same wing.
As a result, the question is no longer really a question.

The forward top part of a cambered wing will always create upwash if it is lifting.  Momentum (mass of air molecules) plays a key role in creating the lifting low pressure area above the wing.
Reynolds number also plays a key role, particularly velocity.  "Top lift" can  only be created if velocity is sufficient (literally) for the wing chord to pass before the low pressure area collapses$^1$.  This tendency for the air to "bend" back towards the wing is the downwash.

if only the front top of the wing pushes air up, and the rest of the top is bending air down, the net effect is lift$^2$ $^3$

Lower Reynolds number wings cannot create top lift, and must rely on "surfing" the air (action/reaction) off the bottom of the wing.  These wings will not create lift at negative angles of attack.
Top lift has a much lower drag penalty.  This is why thin undercambered wings fell out of favor as airspeeds of early planes began to exceed 100 km/hr.
$^1$ angle of attack is also a critical factor
$^2$ this can happen even at a slightly negative AoA.
 $^3$ similar to leverage, the longer distance/time, of downwash zone creates more lifting force than the upwash takes away

Don't know if you're still on this, but I think that your doubt arises from a misunderstanding in terms: the angle of attack we are normally discussing about is the geometric one, i e. the AoA which is referred to the line connecting the leading edge with the trailing edge of the airfoil. This line has no real aerodynamic meaning (unless the airfoil is symmetrical: in that case it coincides with the line of zero lift). The AoA that separates positive lift and negative lift is the aerodynamic one and it's measured in respect to the line of zero lift.
For a not symmetrical airfoil (aka cambered airfoil) the line of zero lift is normally negative, just like in your example:

think about cambered airfoils, some of them have a zero lift angle at around -5°, so if you held one at -3°, you would have lift but negative AoA

Yes, you would have lift at negative geometric AoA but positive aerodynamic AoA.
The geometric AoA is normally used because the line connecting leading and trailing edge is easy to draw and not mistakable, even if that can lead to an airfoil producing positive lift at negative AoA: but that's not against physics, it's just a convention.

what happens to circulation around a cambered airfoil at a small, negative angle of attack which is producing lift, but is also producing upwash

Hopefully now it's clear that a positive lift is produced if and only if the circulation is positive too (and viceversa). That the (geometric) AoA is negative has nothing to do with the sign of the circulation and of the lift.
By the way, since we are dealing here with potential flow around a 2D airfoil, please note that the net downwash around the airfoil is always zero.

Thanks, I do still check this. I get that geometric AoA isn’t a very useful measure, and aerodynamic AoA is more relevant so let me rephrase. Is it possible to have an airfoil at positive aerodynamic AoA, producing upwash past the trailing edge. This appears to be what ‘On Wings of the Minimum Induced Drag: Spanload Implications for Aircraft and Birds’ suggests.

@AeroDude12: wait, you're mixing up airfoil with wing. Airfoil is a 2D shape which has no (or infinite) span. The only circulation involved is the one that you see for example in the picture of this answer. A wing is a 3D object with a much more complex circulation environment, as visible for example in this picture from this answer. There you see that all the vortices shed from the trailing edge coalesce in two vortices, one behind each wingtip...

... Those two vortices (called tip vortices) have an upward component which is used by birds when the fly in formation.

I understand gamma and circulation in 2D aerodynamics, and the kelvin theory leading to both wingtip and starting vortices. But it is unlikely to be the case that the wingtip vortex is what birds use. If you look at the studies done on it they often have to appeal either to incorrect flying by birds or wake contraction (which is totally unobserved) to explain why they appear to fly in a different region than predicted with greater standard deviation. The reason, as outlined in the above paper, freely available on the NTRS server, and by a thesis by Lukacovic is upwash before the wingtip.

https://ntrs.nasa.gov/api/citations/20160003578/downloads/20160003578.pdf Here is the NASA paper by Bowers et Al. https://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article=3631&context=theses And that’s the thesis (it’s very long I’m sorry).

@AeroDude12: I did read the (very interesting indeed) NASA report but, still, your question is about 2D airfoil and for 2D airfoil the sign of the circulation and the one of the lift just match: $\Gamma>0 \Leftrightarrow L>0$. Now, if we go in 3D, then also the spanwise variation of lift and circulation becomes important and it can give rise to "strange" phenomenon like the one you read in the paper where an upwash is $\underline{locally}$ possible; anyway $\underline{globally}$ the net downwash is always negative (if lift is positive) in order to obey the law of conservation of momentum.

Right? So how is global downwash measured bs local downwash?

@AeroDude12: I'm pretty sure the NASA paper describes how the measurements were done.


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Why did my flight leave the gear down for the first 10 minutes of flight?
This morning, I took flight VS156 from LAS to LHR, aboard a 787, tail number G-VSPY according to flightradar24. The pilot made a strange announcement before takeoff - specifically, that due to a recent maintenance procedure, the pilots would not be able to raise the gear for the first 10 minutes of flight. The rest of the flight continued normally. What was the procedure likely to be, and why would leaving the gear down be safer?

I have not heard of that situation occurring for maintenance reasons, but I know that it is done occasionally when the brakes are too warm due to a previous flight being on the heavier side and a really quick turnaround time. Leaving the gear down in that situation would be giving it the cooling that it did not get at the gate.

That would be quite surprising here, considering that the turnaround was around four hours (and the weather was fairly cool)?

@MikeSowsun Nice find - I suspect this answer to that question is most likely to be the case here. Or 757toga's answer below.

@757toga What's suggested in the other answer I linked to would be an MEL for a particular wheel's brake, with a proviso for time with the gear extended to allow the wheel to spin down (without the normal braking) before retracting it. The other answers to that question seemed unlikely to me too, because they do seem to beg the question of getting maintenance done before flight. I don't know if an MEL such as that answer posits does exist for the 787; the reference you provided makes the fire detect MEL a likely scenario. Either way, engine-out + gear-down climb data would clearly be needed.

What's weird to me is why would they tell you about this?  I'm sure random stuff like that happens all the time, but the passengers really don't have any need to know about it, and telling them can only lead to confusion or possibly even panic, which is the last thing you want.

This was to reassure us that any additional noise/buffeting in the cabin was normal.


text:
Currency requirement for a Private Pilot adding a glider rating?
I have a Private Pilot certificate {ASEL, instrument) that I got years ago, and I haven't flown airplanes in over ten years. I'm trying to get back into flying now, but not in airplanes; I want to get a glider rating.
A couple folks I spoke to said that there's an unfortunate quirk in the FARs, where I am required to be current -- that is, have a flight review -- before I can get my glider rating. This would entail spending some serious time and money, working with an instructor to get proficient enough in airplanes again, and get signed off.
Of course, this feels like a step in the wrong direction for me; I want to fly gliders, not airplanes! Things would be simpler if I didn't have my PPL at all, and just started as a student pilot in gliders...
The people I chatted with are knowledgeable -- one is a very senior instructor, the other a DPE -- and they don't have anything to gain by telling me I need a BFR, so I do trust them. But I still want to ask: is this BFR really necessary? What are the relevant FARs that dictate this? Is there any reasonable way around it?

Why don't we get this changed?  AOPA, are you listening?

A Private Pilot certificate includes Sport Pilot privileges. The ubiquitous SGS 2-33 is a sport-legal glider*. You may use the Sport process to train with one instructor and demonstrate proficiency with a second instructor. There's some paperwork to send off, but the "checkride" with the second instructor counts as a flight review (Please don't say BFR, it's so dated, deprecated, wrong).
*There are several other sport-legal gliders, but I don't recall them off the top of my head. The lovely ASK-21 is not sport-legal.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.

As another answer has said, you cannot just solo without being current. You will need to have a flight review, or something that counts as a flight review.
Adding another rating is most likely out of the question, but completing a phase of WINGS, an FAA pilot proficiency program, also counts. This involves 3 earning "knowledge credits" and 3 "flying credits" on three different topics. Critically, it seems to be possible to earn the flying credits in a glider, so you should be able to complete a phase of WINGS and become current without having to touch an airplane.
Alternatively, you can surrender your pilot certificates and "start fresh" with glider training. You would then need to take the written as well, I believe.

Unfortunately, one of the three required WINGS basic-level credits for gliders is a glider flight review. 61.56(c)(1) implies that you can't have a flight review in a category of airplane you're not rated for, so sadly I don't think the WINGS strategy works here :(

@mergeconflict There are many different options for WINGS credits. I haven't looked through all of them, but I think some should be possible.

You're absolutely right, I didn't realize there were other options than the defaults! One option is called, very appropriately, "Glider Student Activity." Best of all, this activity provides all three credits for both the basic knowledge and basic flight requirements!


text:
Whats the difference between "Type Certificate" and "Type Certificate Data Sheet" and how it fits into airworthiness
From my knowledge, a type certificate is given to the manufacturer which in turn allows them to get an airworthiness certificate for the plane it manufactures without having to go back to the FAA to ask for it. I know the TCDS outlines the general information about the design and assigns specific numbers to it. That is about all I know about TCDS though. If anyone can tell me more about the two and how they relate to each other and airworthiness it would be greatly appreciated!

The Certificate Data Sheet is a component of the "Type Certificate"

§ 21.41 Type certificate.
Each type certificate is considered to include the type design, the
operating limitations, the certificate data sheet, the applicable
regulations of this subchapter with which the FAA records compliance,
and any other conditions or limitations prescribed for the product in
this subchapter.

In other words if Boeing holds the Type Certificate for the 737 they hold all the above listed documents including the Certificate Data Sheet, this in turn allows them to apply for "Production Under Type Certificate" to actually build the airplane. If you want to see whats included in the TCDS, they are public and you can check them out here for your favorite airframes.


text:
How does drag change efficient cruise conditions?
If I have an aircraft, and I reduce the induced drag produced by it, does this mean I should operate it at a higher altitude and faster speed to take advantage of the drop in total drag, or slower at lower altitude to take advantage of the lower green dot drag?

It depends on the numbers. "reduce the induced drag" is vague. And we know nothing about the engines (altitude matter). What about same altitude, same speed, less fuel consumption? Try with real numbers.

@sophit What is an airliner's "green dot speed", and how does it vary with weight, and why?

@Bianfable: thanks, never heard about it

Would it be accurate to say that you specifically mean you reduced the coefficient of induced drag?

Thanks people! This is all extremely helpful and really useful!

The easiest way to reduce induced drag in an aircraft is to make it lighter.  However, I'll assume you're talking about design / configuration changes, not operational ones.
If the drag polar has the simple form:
$C_D=C_{D,0}+\frac{C_L}{\pi\,e\,AR}$
We can talk about the balance of induced and parasite drag.
If you change the balance such that induced drag coefficient goes down (say by improving $e$), you'll need to fly at higher lift coefficient to achieve best L/D.
$C_L=\frac{W}{q\S}$
$q=0.5\,\rho\,V^2$
So, if wing loading stays the same, to fly at higher $C_L$, you want to fly at lower $q$.  This requires flying at lower speed or higher altitude.  You probably want to maintain your cruise Mach number, so you'll fly at higher altitude.
This helps explain why the Global Hawk flies where it does.

Thank you, very helpful!

We use the usual approximation that the drag coefficient for the entire airplane can be expressed via the well known:
$C_d=C_{d_0}+kC²_l$
The first term represents the "parasite drag" while the second term is the "induced drag".
Equating the power requested to win that drag $C_d$ with the power available from the engine(s), the results of the following table are obtained:

Speed best range
Speed best loiter

Propeller
$\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{d_0}}}}$
$\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{3C_{d_0}}}}$

Jet
$\sqrt{\frac{2W}{\rho S}\sqrt{\frac{3K}{C_{d_0}}}}$
$\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{d_0}}}}$

Being $K$ at the numerator, all the rest being the same a reduction in the induced drag always implies a reduction of the optimal speed (of whatever you'd like to optimise, either range or loiter).
If you want to compensate for this reduction and increase again the optimal speed, then you have either to fly at a lower density $\rho$ (i.e. higher altitude) and/or increase the wing loading $W/S$ and/or reduce the parasite drag $C_{d_0}$ as well.

The other assumptions commonly left out for these 'best' conditions include: fuel consumption (BSFC or TSFC), propeller efficiency, and drag polar all constant with flight condition -- airspeed / Mach, altitude, throttle.  In reality, none of these things are true.  In practice, they are close enough as long as you're careful.
The variation of max L/D with Mach number sets the cruise speed for a transport.
The variation of TSFC with altitude TSFC=TSFC$_0\sqrt{\theta}$ drives transports to want to cruise at the Tropopause (11km, 36,089ft).

@RobMcDonald: well, there are other simplification too other than what you've listed (a real polar has three terms at least; the weight changes during the flight; and so on). Anyway those simplified equations are enough to grasp the basic behaviour of an airplane in terms of speed optimisation. And... this is not Journal of Fluid Dynamics  Thanks anyway for you comment, which applies equally well to your own answer

That is why I left my answer at CL and best L/D.  I was trying to stray away from the specificity that leads to needing the rest of the complexity.
Weight change during flight is captured by your equations, you aren't ignoring anything there.
The simplicity of the drag polar was clearly stated.  The fact that you're solving for a best speed -- using information that in reality varies with Mach number is a hidden nuance.

@RobMcDonald: if it's ok for you I'll expand my answer with you suggestions

By all means, go ahead.

Thanks people!!

You may notice airliners are designed to have their Vbg (best glide) very close to their desired cruising indicated airspeeds.
It is better to go high, because at higher flight altitudes True Airspeed is greater then Indicated Airspeed.
This means the plane goes farther for the same amount of "push".
All well and good until True Airspeed gets too close to Mach 1, the speed of sound.  Then drag begins to rise.
So you want to fly "Green dot" indicated as high as you can until Mach number increases to the point where fuel economy is lost.
Subsonic airliners use swept, supercritical wings to get as close as they can to Mach 1 economically.

Thank you, although most airliners are not operated at green dot but faster, since their total fuel consumption is lower over a given DISTANCE if you fly faster.

@AlastairWyllie lowest fuel consumption per distance is at Vbg indicated.  Airliners indeed fly faster True Air Speed at higher altitudes up to Mach limits, but for best fuel economy they want to be around Vbg indicated, just like gliders.


text:
Why are there no new major aircraft in ICAO Aerodrome catagory D?
So the 757, 767, A300 and A310 all were in ICAO cat D (mainly restricted by a wingspan between 36-less than 52m). Now though all these aircraft are discontinued, and there is no new designs looking to directly replace them. I’d love to know why this is, whether there are still gates for D class aircraft and whether aircraft such as the A321neo could start to make a business case for Airbus’ replacement for the A320 family to go up an ICAO catagory in exchange for a big fuel efficiency improvement?

If the question revolves around "why" the answer is probably "money."

Don't know for sure, but I'd say as Zeiss says, it's getting harder and harder to raise capital for that kind of project. The C Series nearly bankrupted Bombardier, and they were forced to give it away to Airbus for free.  Ever since the 2008 crisis, banks have been much more risk averse than before it and risk capital has been harder to raise.

That’s a good (if slightly depressing) point. I guess Airbus will do a lot of cringing over any new designs from now on. Of course, once you commit to making a new design you are already taking a massive capital risk


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What on Earth is Going on with NASA's PRANDTL-3 Wingtips?
I'm afraid this will require some explanation. I'm sorry, this will be a long read but I hope it's an interesting conundrum. If anyone reading this isn't familiar with NASA's PRANDTL wing research, Al Bowers' main paper is here: https://ntrs.nasa.gov/citations/20160003578
As a brief summary, Bowers and his team surmise that a Bell-Shaped Lift Distribution (BSLD) rather than the ubiquitous elliptical one, combined with very pronounced wing twist, can produce 'proverse yaw' which they explain as a portion of the outer wing producing a lift vector tilted forward. They then design a series of gliders based on this theory which are flying wings, with no vertical stabiliser. These gliders are tested, and it is found that they can in fact execute coordinated turns without a rudder or other independent yaw control surface.
Then, in answering this question: Why isn't the bell distribution used? Peter Kämpf's excellent answer suggested that in fact at most angles of attack (I.E. except near stall) the BSLD is actually producing negative lift at the tips, which seemed to agree with the stall analysis done of the design: https://ntrs.nasa.gov/citations/20180006832 in tandem with another paper outlining the model validation tests: https://ntrs.nasa.gov/citations/20210014683 this perfectly well explains the proverse yaw and controllability without any controversy.
That seemed to be that, however, keen to examine the wing further, particularly in the assertion by Bowers and his team that the flying wing would yaw out of a slip angle back into the prevailing wind direction, which I still do not understand, I did some CFD. I downloaded a CAD model from the internet which was shared by the team, stitched it, stuck it into simscale with no slip angle, and no adjustment to angle of attack from the CAD file, and did a few quick check runs to see if it all worked nicely. It did, but the results seem to me pretty startling... my question is, looking at the CFD below, what on earth is going on?! The simulation can be found here in case you wish to interrogate it yourself: https://www.simscale.com/projects/vh23138/prandtl-d_incomp-/ it's an incompressible simulation with potential flow initialisation at 13.7m/s flight velocity (45ft/s as per above papers).

CAD file tip geometry: this is the CAD file in Fusion 360, the wingtip foil is symmetrical, and at an AoA of around -2.2 to the flow axis

SimScale wake: There is clearly upwash along the outboard region of the wing,

SimScale pressure slices:
a) slice taken just above the port wingtip, location shown in second image,

b) slice taken just below the port wingtip, location shown in second image

SimScale wingtip pressure distribution: this is a slice taken as a cross-section through the wingtip airfoil, showing a sort of chord wise pressure distribution as far outboard on the starboard wing as I could get without there being virtually no airfoil there, the location is shown in the second image (sorry, it's hard to see the wing wireframe).

So my question: 1. shows symmetrical airfoil at -AoA, 2. shows upwash at that foil, 3. appears to show modest lift production, or at least, a marked lack of clear downforce. Which leads to the question: How?! currently the best I can assume is this a 3D aerodynamic effect produced by the flow field around the whole wing, but that isn't an answer...

Having done some more thinking and a little more investigation of the flow around the wingtips in SimScale, I think I have a theory which explains the observations. The important thing to note is the angle of attack experienced by the wingtip is not the same as the relative angle of a given tip airfoil to the horizontal. Examining again the CFD, this can be clearly seen using streamlines, which appear to be close to the same angle of attack as the wingtip, and so the wingtips could conceivably be at a slightly positive incidence (see below, Red line is added on afterwards, showing chord line of top airfoil)

Now the question has become 'Why is the airflow experiencing such pronounced upwash before the tips?' and this I think can be explained in a pretty standard way: the wing has a 24 degree sweep angle, and it is known that swept wings experience a pronounced reduction in downwash toward the tips, which can sometimes become an upwash, due to the influence of the inboard (and more forward) portion of the wing's trailing vorticity.
This is however, only a working theory, and I have no real way of testing it without redesigning the PRANDTL-3 wing with varying sweep angles and observing wingtip flow behaviour. Any answers with more concrete explanations would be welcome...

This isn't really an answer and should be appended to your question with an edit.

I agree that it isn’t using a hugely rigorous evidence base to answer the question, I’m just looking at relative airflow angles, but the question is not hugely rigorous. This ‘answer’ is in no way clarifying, adding to or correcting the question, it is attempting to posit an explanation. As a result, I felt it wouldn’t be an appropriate addition to the question since it attempts to answer it


text:
Wingwalking at cruise on a jetliner
I have always wondered what the minimal equipment and best strategies would be for an average person to survive as much time as possible in cruise on a standard jetliner. Assume temperature is 20 c.
Would either being at the end or near the body have any effect on time to death? Could they survive by simply holding on? Would they need any form of rope to restrain them?

"Assume temperature is 20 c" - well, that's a problem, because it's not. The temperature at jet cruise level is very, very cold. About -54 °C at 35,000 feet. See What is the typical temperature of an airliner's hull during flight? for more discussion.

Best strategy: stay inside.

You'd get blown off, pass out shortly after, die from hypoxia and/or hypothermia shortly after that, freeze, then hit the ground partially thawed.

Not if you have someone holding you by the belt...https://en.wikipedia.org/wiki/British_Airways_Flight_5390

A heated suit with helmet, sealed oxygen mask and a beefy restraint.

Heated: it's cold up above 30,000 ft, usually below -50°C.
Helmet and oxygen mask: he static pressure is still enough if breathing pure oxygen, but at minimum a helmet and sealed mask are needed to be able to face the air stream.
Beefy restraint: terminal speed is ~100 knots. That's when the drag (of human facing the stream as jumpers normally do) balances the weight. Airliner is doing at least 250 knots indicated and 2.5 times the indicated speed means over 6.25 times the dynamic pressure.


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Why is my calculated seat-miles per gallon of small aircraft higher than of big aircraft?
I've calculated the seat-miles per gallon for three planes. This factors in the speed; it basically means how much does it cost to transport 1 person 1 mile.

Plane
Seats
Cruise (MPH)
gph
seat-mpg

Cherokee 6
5
160
16
50

Grand Caravan EX
14
207
65
44

Rutan Quickie Q2
2
155
3.6
86

This is really surprising to me; I would have expected the exact opposite to be the case. I would have thought that as you group more cabin space together, you get more efficient.
If true, why do bigger planes get proportionally less efficient and more expensive per seat mile?
But if my math is right, and you want to move 14 people from point A to point B, you'll use less fuel sending them in 3 Cherokee 6's than in 1 Grand Caravan. I picked the Caravan because it appears to be the most efficient one out there; any of the other turboprops or twins are far worse. What's going on here?
On the road, the bigger the vehicle the more efficient (buses for example). What's different about the air?

You should compare at the same speed.

@mins my calculation is gallons per seat-mile, which factors in speed. It basically means, how much does it cost to transport 1 person 1 mile.

@IsaiahTaylor Drag goes up with the square of speed, so all else being equal, fuel per hour goes up faster than miles per hour, so fuel per mile goes up as well.

Gallons per seat mile does not take into account speed. It takes into account miles, seats and gallons.

And, you are coparing apples to oranges. The planes you list are very different conceptually, so the really don't tell squat about what you are asking for. You should compare different variants of the same plane family, like A320 vs A340. And that is a bit far fetched too...

Seat-Miles is not the only consideration. 1) Remove one seat per aircraft for the pilot. Recalculate. 2) Operating a 15 seat aircraft is not 3x the cost of a 5 seat.

How many times would you need to overhaul the 3 Cherokee engines before the PT6 of the Caravan?  There are way too many tangible and intangible factors to this question...

All else being equal, your intuition is correct that a larger aircraft will be more efficient (in a gallons per seat-mile measure) than a smaller aircraft.  That's why there are big aircraft like 747's, 777's, and the A-380 flying on routes that have enough traffic to support them.  When you have 300 passengers, it's more efficient to put them into one aircraft that holds that many than into two 737's to fly them the same distance.
Similarly, when all else is held constant, it's more efficient to go slowly than to go fast... how much effort does it take to pedal your bike at 5 mph vs at 20 mph?  If you don't have the stopping and starting of city driving, your car will get better mileage at 30 mph than at 70 mph (assuming the transmission keeps the engine in an efficient range at both speeds) -- you're spending less energy overcoming wind resistance.
And the same "all else held equal" statement can be made about heavier vs lighter aircraft.  For the same aircraft at the same speed & the same altitude (and winds, and temperature, and etc), the heavier one will use more gas to get where it's going.
The problem with the computations is that all three of these variables are changing when hopping between the Rutan EZ to a Cherokee 6 to a Caravan... speed is changing, the aircraft weight is changing, and capacity is changing.  When all of these are changing, it's no longer a case of "all else being equal" and it's no longer valid to draw conclusions about the relationship of one of the variables (capacity, in this case) to efficiency.
If the comparison were between a Cherokee 6 and a 4-seat Cherokee cruising at the same speed, then you'd see results matching intuition.
Put another way, the fact that a 2-seat moped is more efficient (in gal/person-mile) than a 4-seat sedan is, doesn't mean that a 60-seat bus won't be more efficient than the sedan; it means that everything else that changed going from the moped to the sedan, obliterating the "all else being equal" assumption, overwhelms the inherent efficiency of the bigger vehicle.  And, in many cases, we happily accept the increased person-mile costs of the sedan, since it is faster & safer & a more pleasant ride than the moped.
Same deal with the Caravan... it's faster, it's more powerful, it's a lot of things that make it more desirable than the Cherokee, and those not-all-else-being-equal factors make it look less efficient in this case.  But that's more indicative of the distinct nature of the two aircraft, than information about the general trend of more seats making for better efficiency.
(Also, unless fuel were your only cost -- pilots & maintenance are free -- you wouldn't use 3 Cherokee flights to move the 14 people... all the other costs would make the Caravan a better deal financially, even if speed weren't a factor in the decision.)

This old question addresses one of the specific differences:  https://aviation.stackexchange.com/questions/29588/are-turboprops-more-efficient-than-piston-engines-thrust-per-fuel-consumption

@Adam Nice find -- turboprops are preferred for lots of applications as being more economical to operate than piston engines overall, but not in the narrow definition of fuel consumption. By the time the operator factors in the bills for maintenance and the benefits of lighter, less complex engines, the fuel bill is just one of many factors.

Thanks so much for this answer, it clarified a lot. It seems to me that the main thing is that the big planes go faster, so they cost more. If someone built a slow-flying Caravan, the cost might scale similar to a Cherokee 6.


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Do aircraft travel faster at higher altitudes?
I'm aware of the difference between TAS and IAS, and how IAS gets lower in reference to TAS as you climb and the air gets thinner. What I'm trying to figure out is whether or not an aircraft's IAS stays somewhat constant as it flies higher, translating to an increase in TAS.
Do you get a benefit in TAS at FL250, or is your TAS similar to what your TAS was at 3000ft, and now your IAS is much lower?
Does this answer change if it's jet or turboprop or turbocharged piston or piston?


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How to correct the design of an aircraft that is not speed-stable during approach?
I was just wondering if an aircraft is found to NOT be speed stable during the approach flight phase, what might a designer be able to do to stabilize the aircraft without adversely impacting its landing performance?
Using airfoils with reflex camber or the use of a combination of sweep and geometric twist should help but won't that lead to compromising of aerodynamic performance?

"During approach" suggests that setting flaps for approach changes what had been stable before to unstable. Could you please explain the configuration in more detail?

It sounds as if the design process has already gone through manufacture of the aeroplane and through flight testing, which is a very late stage to re-design the complete main wing geometry and structure. There are some items open to interpretation in your question, let's consider two cases:
1. Stick fixed static stability

Assuming that the design in question is not a tailless one, the best approach is to work on the secondary control parameters:

tail volume - increase horizontal tail area $S_h$ or tail length $l_h$;
stabiliser incidence angle - increase maximum deflection angle as it may be in the wake of the fuselage or main wing;
flaps - deflecting them results in a nose-down $C_m$

After some simplifying assumptions, the moment equation in glide becomes:
$$C_m = C_{m_{ac}} + C_{N_W} \cdot \frac{x_{cg} - x_W}{\bar{c}} - C_{N_h} {\left( \frac{V_h}{V} \right)}^2 \frac{S_h \cdot l_h}{S \cdot \bar{c}} = 0$$
So the three variables that can be adjusted are

$C_{m,ac}$
location of the centre of gravity
tail volume $S_h \cdot l_h$

For stability, $dC_m/d\alpha$ < 0. The graph above shows the contributions to $dC_m/d\alpha$ from wind tunnel measurements on a Fokker 27 model. The (unswept) wing has a small contribution, introducing wing sweep and twist would help stability - but would be totally out of context for the F27 mission. The main contribution to stability of the model is from the horizontal tail.
2. Back end of the power curve
Image source
At the back end of the power curve ($V_1$ point A in above graph), there is more power required when the aircraft slows down. The airspeed does not automatically self-correct after a disturbance from a horizontal wind gust, but requires constant monitoring and additional power inputs.
The solution: more available power at low speeds, or a lower wing loading. At high wing loading the $C_L$ of the wing is high, causing the steep rise at the left side of the Power Required curve. Flaps increase $C_L$, and Fowler flaps also expand the wing area during approach.

Why mess with the whole configuration when the issue is wrong hinge moments? Add those hinge moments to your equations and better ways of improving speed stability will become apparent.

That’s not how I read the question!

Yes, I assumed as much. That is why I started my answer with a clarification. Speed stability is clearly defined but it is impossible to know whether the question really means what it says.

If I understand the question correctly, the airplane will exhibit a negative stick force gradient with speed. In other words, when speed goes up the stick will move in "pull" direction when left to itself when for a stable airplane it should move in "push" direction.
From the little your question conveys I come to those conclusions:

There must be a configuration change when what has been stable before becomes unstable during approach. I suspect that wing flaps have been lowered.
The airplane uses reversible, fully mechanical controls.
When moving the stick, speed will change opposite to what certification requirements allow (see CS 23.173 in this document).

Please note that what follows is based on those conclusions. If you misrepresented the issue, don't blame me if the remedy won't work.
Generally, speed stability can be improved by:

Positive camber on the elevator. This will drive its free floating angle (auswehwinkel) to more negative values and will require a trim force to trim the airplane. If this trim force is speed-independent, say by a spring between pushrods and structure, this trimming will preload the elevator in "push" direction and increasing dynamic pressure will move it in "pull" direction. This means more "pull" force with higher speed, ergo higher speed stability.
Reduction of the hinge moment change over angle of attack. This can be achieved by control horns on the elevator. If done excessively, they will even move the elevator in "push" direction when angle of attack increases, but generally a reduction of the elevator's tendency to follow the angle of attack changes should suffice. If the horn reduces the gradient of stick forces over deflection angle too much, add an anti-servo tab.

Depending on what is causing the loss of speed stability, those measures might not work. Your first task, therefore, is to find out what changes in approach and how that change can be neutralized. I have a suspicion that setting flaps increases flow separation at the wing and that this wake is hitting the elevator when flying at approach speeds. Here the best remedy is of course to move the stabilizer up and out of the wing wake.

Using airfoils with reflex camber or the use of a combination of sweep and geometric twist should help

No, not at all. Both will affect performance negatively and have much more impact on other parameters besides speed stability. Your issue is with the change in elevator hinge moment over speed, and this can best be addressed by tailoring the hinge moments properly.

You're describing stick force stability.

@Koyovis ... which is proportional to stick-free stability in a reversible control system. Yes. And what I propose to do should help.

Not speed stable on approach...

Strongly infers that adding flaps moves the CP of the wing so far back that the wing nose down torque becomes stronger than the tail nose up torque, and the stabiliator cannot compensate.
The quickest fix is to re-evaluate the forward CG limits graph with flaps extended.  Moving the limit aft will increase safety margin.
The suggestion of increasing camber of the horizontal stabilizer (rather than size) seemed out for large airliners (due to increased drag at higher Mach values) until ... putting slats and flaps on the stabilator too?
Yes, that would be funny, so increasing tail volume would be a potential solution, with increasing camber a very viable solution for lower speed aircraft.
Thirdly, evaluation of changes in airflow caused by lowering flaps, remembering that the tail creates downforce, so downwash from wing will increase its AoA.  There is a real danger that an (excessively small) overtaxed stabilator could actually stall before the wing, which would be disastrous. Worse yet, a computer relying on limited sensory input might respond by trying to deflect the stabilator even further.


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How can I know the manufacturer of the PFD (Primary Flight Display) of the current mainstream passenger aircraft?
I'm developing a computer vision software based on the photos of these PFD (Primary Flight Display). How can I know the manufacturer of these PFD of the current mainstream passenger aircraft: B737. B747, B777, A320, A330, A340?
I have tried to Google and to find documents about these aircraft, for example, I know the device reference of the PFD of B737 is G1000 made by Garmin, but for others, it's really difficult to find out the device reference of their PFD.
Here is an example of the PFD of B737.

The 737 does NOT use a G1000... Many of these devices are made by companies like honeywell or the makers of the airframe themselves although they may very well source the physical screen from another manufacturer.

Really thanks sir, from the document you provided, does 737 use the G5 EIS of Garmin? Or 737 is not on the list, so that means it does not use a G1000?

I am confused because the list only showed those who installed G5 EIS of Garmin, but 737 is not on the list. I don't know how you know that 737 does not use a G1000 from this list. Can you explain a little bit for me? Really thx!

@StoneSHI the G1000 is not certified for use with the 737 airframe (does not appear on the list) and thus would not be legal to install one (as per the FAA, other jurisdictions may have different rules but most are fairly in line)

The original A300-A320 system was by Aerospatiale, which was part of the Airbus Industrie consortium. Through a series of spin-offs and acquisitions, it became part of an independent supplier, Thales, which ended up with the contracts for all Airbus models.

@user71659 You should add that as an independent answer (ideally with a short quote from the site you linked).


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Using a spring balance to find static EDF Thrust
I have been attempting to measure the static thrust of my EDF with a spring balance. It is part of a group university project. The EDF specs are as follows: 90mm freewing 9 blade 1900 Kv thats rated at 4000g (https://www.rc-castle.com/index.php?...roduct_id=8798). It is being powered by a 6s lipo battery. I made a wooden stand to screw the EDF on. The stand is then attached to the metal sliders. I attached a spring balance to the wooden stand. So when the EDF turns on, the stand slides foward. The spring balance is pulled and the value can be read off it. However, I am unsure if this method is viable as the spring balance is marked incrementally in Kg. I can see this being useful if I hung masses vertically off the spring balance and wanted to measure it, but in my case, I'm trying to measure the force ''horizontally'', if that makes sense.
In summary, I was not sure if I was using the spring balance correctly. Any advice is appreciated.
On the side here are some test results: The EDF at full throttle pulled the spring up to the 3.1kg mark. The lipo was fully charged. I have attached some pictures of it:   . I assume the friction between the rails is responsible for the lower thrust value.

Please note that the 'kilogram-force' or kilopond is out of date. You should use newton... 1 kilopond = 9,8 newton

@xxavier Thanks, will do so moving forward. Unfortunately I could only find these types of labelled spring balances

The setup looks ok, you could check it using a weight on a string, attached somewhere on the thrust line of the fan and then run over a fixed pulley; whatever weight you need to produce an indication of 3.1kg on the scale is equal to the fan’s thrust.

In order to cancel friction effects, push the sliding stand forward as if adding thrust while the motor is running, then release and see to what value it moves back. The most credible thrust value is the average for measuring with and without pushing the stand. If your scale has been made for use on Earth (which I assume), the technique is well suited for thrust measurement once the Kilograms are converted into Newtons by multiplying with the Earth's gravitational acceleration.

To get rid of the sled friction you could hang it from above. Using long enough strings or rods would eliminate the effect of the slight rise of the sled as it moves forward. What is the accuracy you need to aim for?

@Frog so hang a 3.1kg mass over a pulley system and if it is lifted via the pulley system it'll confirm the spring balance reading result? Peter Kämpf we actually did that once at full throttle. The spring balance read 3.5 kg when the slide was pushed abit forward and then returned back to 3.1 kg.
Jpe6 I would say a good estimate at least; Its the thrust supplier for our Electric ducted  fan rocket project

@Johan M, not quite, if the weight of a 3.1kg mass gives the same extension on the spring balance as the fan does then it’s fair to say that the fan has a thrust of 3.1kgf (30.4N).

@Frog I see, thanks

As pointed out by @xxavier, the conversion between the indicated kilograms and the desired newtons of force is 9.80;  more exactly, the local acceleration of gravity, depending on your location and planet.
When used as designed, an object is hung on the hook below the scale.  The force of gravity pulls on the object and stretches the spring, moving the pointer.  The amount of motion is proportional to the force applied.
It is the manufacturer of the device who decides that, if a 1 kg mass stretches the spring to a certain point, then that point should be labelled with "1 kg".
There are many problems with such a system.  It is only correct for one particular value of $g$.  The spring extension may change with age, abusive use, or temperature, or because of external tampering.  Such devices are often marked "Not legal for trade", and balance scales, that match an unknown mass against known masses are labelled "Honest weight; no springs".
There is one problem with the setup you are using.  The spring scale is designed to indicate the mass hanging from the scale. The calibration takes into account the constant load of the moving scale and hook (basically the black stuff in the picture of the scale).
If the spring scale is used horizontally, it will read low, by an amount equal to the mass of the moving scale.
Another factor I would look at is any force acting on the sled from the prop wash of the EDF

I did not even consider the spring scale mass itself, thanks! Would it be fair to measure the mass of the spring scale, and then add it to the measured thrust value itself to account for this? Lastly, my concern was that I'm unsure if the way the manufacturer created the kg labelling system is applicable for horizontal measurements.  It would have been ideal if it was labelled in distance. I could work out the spring constant separately and then find the force with less hesitation

Best way to find out is use a 5 kg test weight, mount the scale horizontally, and attach the rope over a wheel so that the weight hangs down. Use more test weights to calibrate the scale a bit.

@koyovis I initially vertically hung 900g masses on the spring balance hook and it did give a reasonably accurate reading btw

900g is not enough to calibrate the friction and inclination out of this elementary inexpensive spring device.

@Koyovis sorry, can you explain what calibrating for friction means? On a side note, do you think the manufacturer marked the balance in kg in such a manner:  F=kx , mg=kx, then m=(kx)/g for various masses

Yes, basic principle of a spring is that it has a spring stiffness $C = \frac{F}{\Delta s}$. So you are measuring a force with a spring balance. The indication is in kg - but it is analogue, the reading error will be much larger than the variation in gravity along the earth's latitude.
The major issue to be verified is: how much internal friction is there in the spring balance, creating hysteresis in the measurement. When set up for gravity measurements, only the measurement in pull direction would reasonably account for internal friction. So only measuring in the pull direction, your indication would be OK-ish.
A more accurate set-up would be with a wheatstone bridge type force transducer. There are pretty inexpensive Arduino type circuits that can drive them nowadays.

thanks for the reply. Yeah it is not the most accurate setup. We initially wanted to use a load cell, but we did not have any working ones at campus regrettably.


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Servo-actuated aerofoil
I intend on calculating the the servo torque required to actuate a 75mmx38mm NACA0012 airfoil. The example at the end of the page in the link https://github.com/build-week/hover-jet/blob/feature/start-design-scripts/design-scripts/jet_vane_design.ipynb mounts the servo at 25% chordlength and then finds the pitching moment. I would like to know how the servo torque required would be found if the servo was offset from this 25% chord location.
Kind regards

At what airspeed?  At what deflection angle?  If this is a vane in a pipe (a jet exhaust), how big and what shape is the pipe?

Hi, airspeed: 73.27m/s , not sure what exact deflection angle but it is within -15 to 15 degrees , it lies by the exit of a rocket fuselage of 90mm diameter ( it is an electric ducted fan rocket) . We are using vanes to deflect the thrust by the way

See what the L/D curve for NACA 0012 is at 15 degrees AoA, and calculate the Reynolds number for that chord and airspeed.
That estimates the maximum lift, i.e., force at right angles to the duct.  Then you know, by leverage, the max force exerted by the servo on the vane, and from the servo arm's length, the max servo torque.
But that estimate of lift will be corrupted by

the swirly airflow hitting the airfoil
the turbulence between the wall and the "wingtips" that don't quite
reach the wall
the short distance "above and below" the vane to the duct walls, like the middle wing of a Fokker triplane
the interactions at the duct's exit
dynamic effects, if the vane's AoA changes quickly

So for a trustworthy answer, you'd have to do some bench measurements, perhaps with smaller ducts and slower efflux.

Thank you for the informative response! I have actually been playing with Xflr5 to get coefficient values at the Reynolds number. Just a query of the 1st statement: One could find the lift coefficient and find its value with a chosen vane area. Though, I am not certain how one would find the centre of pressure at 15 degrees. Would it be applicable to find the coefficient of moment and use that to size the torque required for the servo alternatively?

Sure, that would work.  But to really avoid servo blowback for even a fraction of a second, use a safety factor of two, because it could get really turbulent in there.  Any by-the-book corrections for max torque (from, say, how CP varies with AoA) of +-10% would get lost in that noise.

thanks for the confirmation